Most proofs of the characteristic polynomial of the companion matrix–an important specific case–proceed by induction, and start with a matrix. It strikes me that an inductive proof has more force (or at least makes more sense) if a larger matrix is used. In this case we will use a “large” (numerical analysts will laugh at this characterisation) matrix.

Let us begin by making a notation change. Consider the general polynomial

For this to be monic (one of the requirements for the polynomial in question) we should divide by the last coefficient, thus

Our object is thus to prove that this (or a variation of this, as we will see) is the characteristic polynomial of

The characteristic polynomial of this is the determinant of the following:

(For another application of the characteristic polynomial and the companion matrix, click here.)

To find the determinant, we expand along the first row. But then we discover that only two minors that matter: the one in the upper left corner and the one in the upper right. Breaking this up into minors and cofactors yields the following:

The second matrix, however, is an upper triangular matrix with ones for all of its diagonal entries. Its determinant, therefore, is unity. Also rewriting the coefficient of the second term, we have

or

Repeating this process for the next set of minors and cofactors yields

Note carefully the inclusion of in the second term. We can also write this as

Repeating this process until the end, it is easy to see that

or more generally

where is the degree of the polynomial (and the size of the companion matrix.) If we drop the terms we used to make the polynomial monic, we have at last

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Interessting thoughts

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